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Let $A(S)$ denotes the Arc complex of a finite type hyperbolic surface $S$ with nonempty boundary. Let $\lambda:A(S)\rightarrow A(S)$ be a map such that on triangulations of $S$ i.e. on the top dimensional simplices of $A(S)$ the map $\lambda$ is induced by a homeomorphism $\Phi$ of $S$.

Question: How to show that $\lambda$ is induced by a homeomorphism of $S$ in the whole of $A(S)$?

P.S: In the context I have heard about it the speaker said it followed from "Mosher's connectivity argument." I searched about it but I didn't find any reference or answer.

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    $\begingroup$ Well, Lee Mosher is active on MO, so no doubt he can explain! $\endgroup$
    – HJRW
    Commented Sep 8, 2013 at 16:48

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I'm not as familiar with this area as I should be, but it seems as though the desired result follows from the main theorem of the following paper:

Irmak, McCarthy. Injective simplicial maps of the arc complex, Turk J Math, 34 (2010) , 339 – 354.

Here is their Theorem:

Let $S$ be a compact, connected, orientable surface of genus $g$ with $b \geq 1$ boundary components. If $\lambda : A(S) \to A(S)$ is an injective simplicial map, then $\lambda$ is induced by a homeomorphism $H : S\to S$.

If $\lambda$ is induced by a homeomorphism on top simplices of $A(S)$, then it is at least (I hope) injective. I see that Mosher's connectivity argument is mentioned and used in this paper as "Theorem 3.9", so presumably this argument is exactly what you're looking for.

Finally, in case you are hunting for more references: I saw all of this in the lovely monograph by McCarthy and Papadopolous called Simplicial actions of mapping class groups. It is available here.

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  • $\begingroup$ Thank you very much for the references. Actually the place I was reading this from takes the opposite approach. It states that if the map is injective on Arc complex then by some geometric property of the map you can show that given any triangulaton i.e a top dimeasional simplex there is a homeomorphism of the surface which induces this map. And then to show that the full map on the arc complex is induced by the homeomorphism it uses Mosher's connectivity argument. $\endgroup$
    – Cusp
    Commented Sep 11, 2013 at 4:51
  • $\begingroup$ Well, at least now you have the statement of the connectivity argument and a reference to it. Good luck! $\endgroup$
    – Vidit Nanda
    Commented Sep 11, 2013 at 13:41
  • $\begingroup$ This is exactly the result I was trying to understand. I realised it after going through the first reference. Thanks again. $\endgroup$
    – Cusp
    Commented Sep 11, 2013 at 16:38

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